2x^2+3x-3.
Ive got to -2±(sqrt)33/4 but dont know where to go from here.
And with the equation 2x=b^2-a^3b I need to make x the subject.
I presume you need to solve $\displaystyle 2x^2+3x-3 = 0$
There are a couple of ways to do this, but I suspect you are looking for the use of the quadratic formula.
Given $\displaystyle ax^2 + bx + c = 0$ then $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
So for $\displaystyle 2x^2+3x-3 = 0$, a = 2, b = 3, c = -3. Thus:
$\displaystyle x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-3)}}{2 \cdot 2}$
$\displaystyle x = \frac{-3 \pm \sqrt{9 + 24}}{4} = \frac{-3 \pm \sqrt{33}}{4}$
For the other problem: $\displaystyle 2x=b^2-a^3b$
I don't know what you mean by "make x the subject." I am guessing you mean solve in terms of x?
$\displaystyle 2x=b^2-a^3b$
$\displaystyle b^2 + (-a^3)b + (-2x) = 0$
This is a quadratic in terms of b. So using the quadratic formula again:
$\displaystyle b = \frac{-(-a^3) \pm \sqrt{(a^3)^2 - 4(1)(-2x)}}{2 \cdot 1}$
$\displaystyle b = \frac{a^3 \pm \sqrt{a^6 + 8x}}{2}$
-Dan